Eight
| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 26261 | Accepted: 11490 | Special Judge | ||
Description
The 15-puzzle has been around for over 100 years; even if you don‘t know it by that name, you‘ve seen it. It is constructed with 15 sliding tiles, each with a number from 1 to 15 on it, and all packed into a 4 by 4 frame with one tile missing. Let‘s call the missing tile ‘x‘; the object of the puzzle is to arrange the tiles so that they are ordered as:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 x
where the only legal operation is to exchange ‘x‘ with one of the tiles with which it shares an edge. As an example, the following sequence of moves solves a slightly scrambled puzzle:
1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
5 6 7 8 5 6 7 8 5 6 7 8 5 6 7 8
9 x 10 12 9 10 x 12 9 10 11 12 9 10 11 12
13 14 11 15 13 14 11 15 13 14 x 15 13 14 15 x
r-> d-> r->
The letters in the previous row indicate which neighbor of the ‘x‘ tile is swapped with the ‘x‘ tile at each step; legal values are ‘r‘,‘l‘,‘u‘ and ‘d‘, for right, left, up, and down, respectively.
Not all puzzles can be solved; in 1870, a man named Sam Loyd was famous for distributing an unsolvable version of the puzzle, and
frustrating many people. In fact, all you have to do to make a regular puzzle into an unsolvable one is to swap two tiles (not counting the missing ‘x‘ tile, of course).
In this problem, you will write a program for solving the less well-known 8-puzzle, composed of tiles on a three by three
arrangement.
Input
You will receive a description of a configuration of the 8 puzzle. The description is just a list of the tiles in their initial positions, with the rows listed from top to bottom, and the tiles listed from left to right within a row, where the tiles are represented by numbers 1 to 8, plus ‘x‘. For example, this puzzle
1 2 3 x 4 6 7 5 8
is described by this list:
1 2 3 x 4 6 7 5 8
Output
You will print to standard output either the word ``unsolvable‘‘, if the puzzle has no solution, or a string consisting entirely of the letters ‘r‘, ‘l‘, ‘u‘ and ‘d‘ that describes a series of moves that produce a solution. The string should include no spaces and start at the beginning of the line.
Sample Input
2 3 4 1 5 x 7 6 8
Sample Output
ullddrurdllurdruldr
Source
第一次的代码：
bfs+康拓展开;
略有点粗糙，晚上交一发，直接过了，但是时间不是很理想，明天再改进一下。
| Accepted | 24600K | 63MS | C++ | 3370B | 2015-05-09 00:58:11 |
1 #include<iostream>
2 #include<algorithm>
3 #include<cstdio>
4 #include<cstring>
5 using namespace std;
6
7 int hehe[3][3]={1,2,6,24,120,720,5040,40320,362880};
8 int hehe2[9]={1,2,6,24,120,720,5040,40320,362880};
9 char w[4]={‘r‘,‘l‘,‘u‘,‘d‘};
10 int dx[]={0,0,-1,1};
11 int dy[]={1,-1,0,0};
12 class Matrix
13 {
14 private:
15 int t[3][3];
16 int x0,y0;
17 int d;
18 int last;
19 public:
20 void Set_value(int x,int y,char *s);
21 void Set_x(int tx){x0=tx;}
22 void Set_y(int ty){y0=ty;}
23 void Set_Deal(int a) { d=a; }
24 void Set_Last(int w){last=w;}
25 int Get_Deal() { return d; }
26 int Get_Last(){return last; }
27 int Get_x(){return x0;}
28 int Get_y(){return y0;}
29 void Swap(int x1,int y1,int x2,int y2)
30 {
31 int e=t[x1][y1];
32 t[x1][y1]=t[x2][y2];
33 t[x2][y2]=e;
34 }
35 int Get_value(int x,int y) { return t[x][y]; }
36 int Get_hash();
37 void operator = (Matrix & w)
38 {
39 for(int i=0;i<3;i++)
40 for(int j=0;j<3;j++)
41 t[i][j]=w.t[i][j];
42 x0=w.x0;
43 y0=w.y0;
44 }
45 void Show()
46 {
47 for(int i=0;i<3;i++,cout << endl)
48 for(int j=0;j<3;cout << t[i][j] << " ",j++);
49 }
50 };
51 int Matrix::Get_hash()
52 {
53 int rev=0;
54 for(int i=0;i<3;i++)
55 for(int j=0;j<3;j++)
56 rev+=t[i][j]*hehe[i][j];
57 return rev;
58 }
59 void Matrix::Set_value(int x,int y,char *s)
60 {
61 if(!isdigit(s[0]))
62 {
63 t[x][y]=0;
64 x0=x,y0=y;
65 }
66 else
67 {
68 int len=strlen(s),c=0;
69 for(int i=0;i<len;i++)
70 c=c*10+s[i]-‘0‘;
71 t[x][y]=c;
72 }
73 }
74 int vis[4000000];
75 Matrix Mx[10000000];
76 int Ans=0;
77 void output(int now)
78 {
79 if(Mx[now].Get_Last()==-1) return;
80 else
81 {
82 output(Mx[now].Get_Last());
83 printf("%c",w[Mx[now].Get_Deal()]);
84 }
85 }
86 bool check(int x,int y)
87 {
88 if(x>=0&&x<=2&&y>=0&&y<=2) return true;
89 return false;
90 }
91 void bfs()
92 {
93 memset(vis,0,sizeof(vis));
94 vis[Mx[0].Get_hash()]=1;
95 Mx[0].Set_Deal(-1);
96 Mx[0].Set_Last(-1);
97 int Start=1;
98 int End=0;
99 while(End<Start)
100 {
101 Matrix mx;
102 mx=Mx[End++];
103 int nx=mx.Get_x();
104 int ny=mx.Get_y();
105 for(int i=0;i<4;i++)
106 {
107 int tx=nx+dx[i];
108 int ty=ny+dy[i];
109 if(check(tx,ty))
110 {
111 mx.Swap(nx,ny,tx,ty);
112 mx.Set_x(tx),mx.Set_y(ty);
113 int temp=mx.Get_hash();
114 if(vis[temp]==0)
115 {
116 vis[temp]=1;
117 Mx[Start]=mx;
118 Mx[Start].Set_Deal(i);
119 Mx[Start].Set_Last(End-1);
120 if(temp==Ans)
121 {
122 output(Start);
123 printf("\n");
124 return;
125 }
126 Start++;
127 }
128 mx.Swap(nx,ny,tx,ty);
129 mx.Set_x(nx),mx.Set_y(ny);
130 }
131 }
132 }
133 printf("unsolvable\n");
134 }
135 int main()
136 {
137 for(int i=0;i<8;i++) Ans+=(i+1)*hehe2[i];
138 char temp[10];
139 for(int i=0;i<3;i++)
140 for(int j=0;j<3;j++)
141 {
142 scanf("%s",temp);
143 Mx[0].Set_value(i,j,temp);
144 }
145 bfs();
146 return 0;
147 }
148 //注释下次一起写